Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $22.2$ years; the standard deviation is $2.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $19.9$ years.
Answer: $22.2$ $19.9$ $24.5$ $17.6$ $26.8$ $15.3$ $29.1$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $22.2$ years. We know the standard deviation is $2.3$ years, so one standard deviation below the mean is $19.9$ years and one standard deviation above the mean is $24.5$ years. Two standard deviations below the mean is $17.6$ years and two standard deviations above the mean is $26.8$ years. Three standard deviations below the mean is $15.3$ years and three standard deviations above the mean is $29.1$ years. We are interested in the probability of a tiger living less than $19.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $19.9$ years and the other half $({16\%})$ will live longer than $24.5$ years. The probability of a particular tiger living less than $19.9$ years is ${16\%}$.